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juin [17]
3 years ago
15

Last one! please please help! THANKS!!!

Mathematics
2 answers:
deff fn [24]3 years ago
8 0

Hey! It looks like here, each week the amount is doubled the week before.

$1 - $2 - $4 - $8 - $16 - $32 - $64 - $128

So, we count up all the numbers, and we get 8 different totals. This would mean that on week 8, Kim can buy her bike! Hope this helped :)

RoseWind [281]3 years ago
5 0

Answer:

week 8

Step-by-step explanation:

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3 years ago
it costs $36.75 for 3 tickets to visit the science center. it costs $51 for 4 tickets to visit the zoo. which attraction will co
lord [1]

Find the cost for one ticket for the science center.

36.75 ÷ 3 = 12.25

Multiply the product with 12 to get the cost of 12 tickets.

12.25 × 12 = 147

Find the cost of 1 zoo ticket.

51 ÷ 4 = 12.75

Find the cost of 12 tickets by multiplying the product and 12

12.75 × 12 = $153

12 Science Center Tickets cost $147

12 zoo tickets cost $153


The <u><em>Zoo</em></u> costs more than the Science Center.


To find out how much more the Zoo costed, subtract 147 from 153.

153 - 147 = 6

The zoo only costs <u><em>$6 </em></u>more for 12 students.

So, you're answers are Zoo and $6

5 0
3 years ago
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To test if the mean IQ of employees in an organization is greater than 100, a sample of 30 employees is taken and the value of t
ivolga24 [154]

Answer:

p_v =P(t_{(29)}>1.42)=0.0831  

For this case the p value calculated is higher than the significance level used of 0.05 so then we have enough evidence to FAIL to reject the null hypothesis and the best conclusion for this case would be:

a) do not reject the null hypothesis and conclude that the mean IQ is not greater than 100

Step-by-step explanation:

Information given

We want to verify if he mean IQ of employees in an organization is greater than 100 , the system of hypothesis would be:  

Null hypothesis:\mu \leq 100  

Alternative hypothesis:\mu > 100  

The statistic for this case is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

The statistic calculated for this case  t_{calc}= 1.42

The degrees of freedom are given by:

df=n-1=30-1=29  

Now we can find the p value using tha laternative hypothesis and we got:

p_v =P(t_{(29)}>1.42)=0.0831  

For this case the p value calculated is higher than the significance level used of 0.05 so then we have enough evidence to FAIL to reject the null hypothesis and the best conclusion for this case would be:

a) do not reject the null hypothesis and conclude that the mean IQ is not greater than 100

5 0
3 years ago
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